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Chemistry: Post your doubts here!

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Muhammad Asif

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smartangel said:
oh God. plz see the question. it says additional isomers. if you switch the positions of CH3 and H it will form a cis trans isomer!!
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miss the question reqiures the isomers other than compound shown of C4H8.......thus there are only three other isomers
 
M

muneexa

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the bond lenghth deosen decrease it increases as the distance b/w the outer shell from nucleus increases thus when ovarlaping takes place the two nuclei gets furthur aapart as we go down the group
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what about bond length in hydrogen halides? does it also increase down the group?
 
nerdybookworm

nerdybookworm

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Ashique said:
For question 2, 2 double bonds in he compound were removed, thus 4 mole of hydrogen was required (since 2 hydrogens are added to the chain once you break a double bond). If you notice, CH2 was changed to CH3 thus one more mole os Hydrogen was required. So in total, 5 mole of hydrogen was required.

Question 4, look for a major change in the ionisation energy. The first 6 value lie close to each other, (or say, one value ot numerically close to the previous value), but the 7th value 13200. Which is BIG increase in ionisation energy. But the first six values lie close to one another. What does this tell about the atom and the ionisation energy? This indicates that 6 electrons were easily removed. And a lot of energy was required to remove the 7th electron. This indicates that the 7th electron lies in a different shell than the first 6. And we can also figure out than the quantum shell (or the outer most shell) has 6 electrons in it. Hence it's in group six. Te is the only element in group VI, hence answe is C.

Question 8- you have to know how to calculate the oxidation number, of a compound in a chemical equations. Go through your book, and I' m sure you'll find out how to do it, (if you don't know that is). The greatets oxidation number change happened in A, which is 20. In B, C and D the oxidation number changes are 6, 4 and 2 respectively.

Question 10- (Since no volume was given, we assume the volume to be 1 dm3, hence the concentrations will have the same numerical value as their mol.)
Now, First calculate the total number of moles at the begining- 0.20+0.15= 0.35 mol.
At equilibrium the number of moles of the products(ie HI) was 0.26. So what is the TOTAL volume of the reactant? 0.35-0.26= 0.09.
Now 0.09 was the TOTAL volume of the reactant. Now look at the denominator, addition of which two concentrations willl give us 0.09? Hence answer is C.

Question 16- Addition of hot alkali will make chlorine go through a disproportionation reaction. Which will give chlorine of two oxidation states -1 and +5, hence answer is D.

Question 20- you simply have to draw out all possible isomers of the compund. It should be 7.

Question 27- You have to find the carbon which has lost the chlorine, since homolytic fission using ultraviolet light, will only give chlorine and the corresponding compound as free radicals. Since C is the only one which lost the chlorine, answer should be C.

Question 28- Reaction of the first three compounds with sodium would give- sodium ethoxide for A and B, with the loss of producing moles of hydrogen, when you balance the equation). Reaction of C with sodium would give sodium ethanote again giving off two moles of hydrogen. It is only compound D which gives of only 1 mole of H2.

Question 30- nucleophilic addiotn of HCN to propanone would give you CH3CH(OH)(CN)CH3. When you reflux this this H2SO4, You'll lose the nitrile and get a carboxylic group , CH3CH(OH)(CO2H)CH3. This compound is represented by D, so thats your answer.
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THankyou soo much,..:)
 
nerdybookworm

nerdybookworm

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Muhammad Asif said:
Q2: see that one chain in residue has 2 double bonds while and 2 chains have single double bonds thus 4 doubl bonds in all....thus only 5 are hydrogenated.....1 double bond needs 1mol H2 so 5 needs 5mol H2

Q4:the highest difference in ionisation energies is 6 and 7th thus the element contain 6 electrons in outer shell thus group 6 that is C

Q8:the only change is in A....

Q10:the molar ratio of HI:I2:H2= 2:1:1 the change in HI is .26 so change in rest would be .26/2=.13......moles at equillibrium would be for H2=.07 and I2=.02 thus the only ans is C

Q16:chlorine disproportionate in conc hot NaOH forming NaClO3 where chlorine is +5(fact)

Q20:there are seven possible structures.....try to draw them

Q27=only chlorine----carbon bond can be broken to form a redical Cl....thus C is d answer as in the other options no Cl bond is broken

Q28=sorry:p

Q30=the answer is D as an additional C is added to Chain and an OH + acid group are made from the product hydrolysis....


q39=fehling test is positive thus the product is an aldehyde which can only be form from a primary alcohol which are shown in only 2 and 3
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thankyouuu soo much..:)
 
furry.panda

furry.panda

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heeelp needed :(

please explain about this:


Flask X contains 1 dm3 of helium at 2 kPa pressure and flask Y contains 2 dm3 of neon at 1 kPa pressure.
If the flasks are connected at constant temperature, what is the final pressure?
A 1 1/3 kPaB 1 1/2 kPaC 1 2/3kPaD 2kPa


ANS is A
 
T

TheMan123

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smartangel said:
i'd be glad if you do :)
Click to expand...
I've read your previous discussion about the question and I see why you don't understand.
The question asks for ANY isomer with C4H8 and with a double bond other than the one shown in the diagram.
The diagram shown is just one example, you can ignore it when drawing other isomers
Like I said before, the 3 isomers are
cis pent-2-ene
trans pent-2-ene
pent-1-ene
Answer is therefore C.3 isomers
 
M

Muhammad Asif

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furry.panda said:
heeelp needed :(

please explain about this:


Flask X contains 1 dm3 of helium at 2 kPa pressure and flask Y contains 2 dm3 of neon at 1 kPa pressure.
If the flasks are connected at constant temperature, what is the final pressure?
A 1 1/3 kPaB 1 1/2 kPaC 1 2/3kPaD 2kPa


ANS is A
Click to expand...

in such reactions the formula is P1V1+P2V2=P3V3
(2*1)+(1*2)=P3*3 -----------> 4k=P3(3)------>P3=4k/3
 
S

smartangel

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TheMan123 said:
I've read your previous discussion about the question and I see why you don't understand.
The question asks for ANY isomer with C4H8 and with a double bond other than the one shown in the diagram.
The diagram shown is just one example, you can ignore it when drawing other isomers
Like I said before, the 3 isomers are
cis pent-2-ene
trans pent-2-ene
pent-1-ene
Answer is therefore C.3 isomers
Click to expand...
it says C4H8..so where the pent 2 ene and pent 1 ene isomers come form? sorry im going a little crazy here :/
 
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